query error

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  • Wm
    #1

    query error

    I have a query that I expect to return 3 or 4 entries -- but I seem to be
    getting only the most recent entry, repeated 4 times. What am I doing wrong
    here?

    $query="SELECT artistID,email, city,state,coun try from artists WHERE
    email='$email'" ;
    $result=mysql_q uery($query) or die(mysql_error ("Could not execute
    query."));
    if (mysql_num_rows ($result) > 0){
    $alreadylisted = "1";
    echo "<CENTER>$e mail is already in our database. Are you listed
    below?</CENTER><BR>";
    while($row = mysql_fetch_arr ay($result)) {
    $artistID = $row['artistID'];
    $email = $row['email'];
    $city = $row['city'];
    $state = $row['state'];
    $country = $row['country'];
    echo "<CENTER><HR><F ONT size=\"2\" face=\"Arial, Helvetica,
    sans-serif\">
    <A HREF=\"javascri pt:;\"
    onClick=\"openP rofile('artist. php?artistID=". $artistID.

    "','Artist','re sizable=yes,wid th=600,height=3 00')\">".$first name."
    ".$lastname ."</A><BR>"
    .$city.", ".$state." ".$country. "</FONT><BR></CENTER>";
    }
    echo "<HR><CENTE R>If you are already listed above,
    congratulations !<BR>
    If needed, you may edit your listing using the link at
    left.</CENTER>";
    mysql_free_resu lt($result);
    }


    Thanx,
    Wm



    Tags: None
    • DjDrakk
      #2
      Re: query error

      try echoing your query to screen so you can read exactly what is being sent
      to the database beforehand.

      --

      Warren Butt
      -- Custom web design, cheap like cheese


      "Wm" <LAshooter@hotm ail.com> wrote in message
      news:1haYa.1802 083$ZC.268155@n ews.easynews.co m...[color=blue]
      > I have a query that I expect to return 3 or 4 entries -- but I seem to be
      > getting only the most recent entry, repeated 4 times. What am I doing[/color]
      wrong[color=blue]
      > here?
      >
      > $query="SELECT artistID,email, city,state,coun try from artists WHERE
      > email='$email'" ;
      > $result=mysql_q uery($query) or die(mysql_error ("Could not execute
      > query."));
      > if (mysql_num_rows ($result) > 0){
      > $alreadylisted = "1";
      > echo "<CENTER>$e mail is already in our database. Are you listed
      > below?</CENTER><BR>";
      > while($row = mysql_fetch_arr ay($result)) {
      > $artistID = $row['artistID'];
      > $email = $row['email'];
      > $city = $row['city'];
      > $state = $row['state'];
      > $country = $row['country'];
      > echo "<CENTER><HR><F ONT size=\"2\" face=\"Arial, Helvetica,
      > sans-serif\">
      > <A HREF=\"javascri pt:;\"
      > onClick=\"openP rofile('artist. php?artistID=". $artistID.
      >
      > "','Artist','re sizable=yes,wid th=600,height=3 00')\">".$first name."
      > ".$lastname ."</A><BR>"
      > .$city.", ".$state." ".$country. "</FONT><BR></CENTER>";
      > }
      > echo "<HR><CENTE R>If you are already listed above,
      > congratulations !<BR>
      > If needed, you may edit your listing using the link at
      > left.</CENTER>";
      > mysql_free_resu lt($result);
      > }
      >
      >
      > Thanx,
      > Wm
      >
      >
      >[/color]


        Comment

        • Wm
          #3
          Re: query error

          "DjDrakk" <DjDrakk@drakkr adio.servemp3.c om> wrote in message
          news:vj2dmnopl9 h08d@corp.super news.com...[color=blue]
          > try echoing your query to screen so you can read exactly what is being[/color]
          sent[color=blue]
          > to the database beforehand.
          >
          > Warren Butt
          > -- Custom web design, cheap like cheese
          >[/color]

          echo "Query result = ".$result;

          yields this output:

          ArrayQuery result = Resource id #21

          I'm not sure where the word "Array" is coming from, nor what the "Resource
          id #21" is.....????

          Wm


          [color=blue]
          >
          > "Wm" <LAshooter@hotm ail.com> wrote in message
          > news:1haYa.1802 083$ZC.268155@n ews.easynews.co m...[color=green]
          > > I have a query that I expect to return 3 or 4 entries -- but I seem to[/color][/color]
          be[color=blue][color=green]
          > > getting only the most recent entry, repeated 4 times. What am I doing[/color]
          > wrong[color=green]
          > > here?
          > >
          > > $query="SELECT artistID,email, city,state,coun try from artists WHERE
          > > email='$email'" ;
          > > $result=mysql_q uery($query) or die(mysql_error ("Could not execute
          > > query."));
          > > if (mysql_num_rows ($result) > 0){
          > > $alreadylisted = "1";
          > > echo "<CENTER>$e mail is already in our database. Are you listed
          > > below?</CENTER><BR>";
          > > while($row = mysql_fetch_arr ay($result)) {
          > > $artistID = $row['artistID'];
          > > $email = $row['email'];
          > > $city = $row['city'];
          > > $state = $row['state'];
          > > $country = $row['country'];
          > > echo "<CENTER><HR><F ONT size=\"2\" face=\"Arial, Helvetica,
          > > sans-serif\">
          > > <A HREF=\"javascri pt:;\"
          > > onClick=\"openP rofile('artist. php?artistID=". $artistID.
          > >
          > > "','Artist','re sizable=yes,wid th=600,height=3 00')\">".$first name."
          > > ".$lastname ."</A><BR>"
          > > .$city.", ".$state." ".$country. "</FONT><BR></CENTER>";
          > > }
          > > echo "<HR><CENTE R>If you are already listed above,
          > > congratulations !<BR>
          > > If needed, you may edit your listing using the link at
          > > left.</CENTER>";
          > > mysql_free_resu lt($result);
          > > }
          > >
          > >
          > > Thanx,
          > > Wm
          > >
          > >
          > >[/color]
          >
          >[/color]


            Comment

            • Wm
              #4
              Re: query error

              "Trent Stith" <trents@airmail .net> wrote in message
              news:bgribl$d9c @library2.airne ws.net...[color=blue]
              > You need to echo the query itself, not the result of the query.
              >
              > echo "<pre>";
              > echo $query;
              >
              > that will return what your query is actually doing it may shed some light.
              >[/color]

              I pasted the lines above in after the query and got this result output to
              the browser:

              "Array
              address@hotmail .com is already in our database. Are you listed below?"

              I'm still not sure why I'm getting the word "Array" output, nor why I don't
              see the query that I'm trying to verify. And I'm still getting the same info
              output 4 times rather than 4 separate listings, but I'm sure that's a
              different problem... <groan>

              Thanx,
              Wm


              *************** *************** *************** *************
              ORIGINAL MESSAGE(S):
              *************** *************** *************** *************[color=blue]
              > "Wm" <LAshooter@hotm ail.com> wrote in message
              > news:oLaYa.1378 547$Ho4.9675373 @news.easynews. com...[color=green]
              > > "DjDrakk" <DjDrakk@drakkr adio.servemp3.c om> wrote in message
              > > news:vj2dmnopl9 h08d@corp.super news.com...[color=darkred]
              > > > try echoing your query to screen so you can read exactly what is being[/color]
              > > sent[color=darkred]
              > > > to the database beforehand.
              > > >
              > > > Warren Butt
              > > > -- Custom web design, cheap like cheese
              > > >[/color]
              > >
              > > echo "Query result = ".$result;
              > >
              > > yields this output:
              > >
              > > ArrayQuery result = Resource id #21
              > >
              > > I'm not sure where the word "Array" is coming from, nor what the[/color][/color]
              "Resource[color=blue][color=green]
              > > id #21" is.....????
              > >
              > > Wm
              > >
              > >
              > >[color=darkred]
              > > >
              > > > "Wm" <LAshooter@hotm ail.com> wrote in message
              > > > news:1haYa.1802 083$ZC.268155@n ews.easynews.co m...
              > > > > I have a query that I expect to return 3 or 4 entries -- but I seem[/color][/color][/color]
              to[color=blue][color=green]
              > > be[color=darkred]
              > > > > getting only the most recent entry, repeated 4 times. What am I[/color][/color][/color]
              doing[color=blue][color=green][color=darkred]
              > > > wrong
              > > > > here?
              > > > >
              > > > > $query="SELECT artistID,email, city,state,coun try from artists[/color][/color][/color]
              WHERE[color=blue][color=green][color=darkred]
              > > > > email='$email'" ;
              > > > > $result=mysql_q uery($query) or die(mysql_error ("Could not execute
              > > > > query."));
              > > > > if (mysql_num_rows ($result) > 0){
              > > > > $alreadylisted = "1";
              > > > > echo "<CENTER>$e mail is already in our database. Are you[/color][/color][/color]
              listed[color=blue][color=green][color=darkred]
              > > > > below?</CENTER><BR>";
              > > > > while($row = mysql_fetch_arr ay($result)) {
              > > > > $artistID = $row['artistID'];
              > > > > $email = $row['email'];
              > > > > $city = $row['city'];
              > > > > $state = $row['state'];
              > > > > $country = $row['country'];
              > > > > echo "<CENTER><HR><F ONT size=\"2\" face=\"Arial,[/color][/color][/color]
              Helvetica,[color=blue][color=green][color=darkred]
              > > > > sans-serif\">
              > > > > <A HREF=\"javascri pt:;\"
              > > > > onClick=\"openP rofile('artist. php?artistID=". $artistID.
              > > > >
              > > > > "','Artist','re sizable=yes,wid th=600,height=3 00')\">".$first name."
              > > > > ".$lastname ."</A><BR>"
              > > > > .$city.", ".$state." ".$country. "</FONT><BR></CENTER>";
              > > > > }
              > > > > echo "<HR><CENTE R>If you are already listed above,
              > > > > congratulations !<BR>
              > > > > If needed, you may edit your listing using the link at
              > > > > left.</CENTER>";
              > > > > mysql_free_resu lt($result);
              > > > > }
              > > > >
              > > > >
              > > > > Thanx,
              > > > > Wm
              > > > >
              > > > >
              > > > >
              > > >
              > > >[/color]
              > >
              > >[/color]
              >
              >[/color]


                Comment

                • James Jiao
                  #5
                  Re: query error

                  Ok insert the echo "$query" line in the following places where indicated
                  with 'echo', that's how you should isolate the problem to one line of code:


                  "Wm" <LAshooter@hotm ail.com> wrote in message
                  news:1haYa.1802 083$ZC.268155@n ews.easynews.co m...[color=blue]
                  > I have a query that I expect to return 3 or 4 entries -- but I seem to be
                  > getting only the most recent entry, repeated 4 times. What am I doing[/color]
                  wrong[color=blue]
                  > here?
                  >[/color]

                  echo
                  [color=blue]
                  > $query="SELECT artistID,email, city,state,coun try from artists WHERE
                  > email='$email'" ;[/color]

                  echo
                  [color=blue]
                  > $result=mysql_q uery($query) or die(mysql_error ("Could not execute
                  > query."));[/color]

                  echo
                  [color=blue]
                  > if (mysql_num_rows ($result) > 0){
                  > $alreadylisted = "1";
                  > echo "<CENTER>$e mail is already in our database. Are you listed
                  > below?</CENTER><BR>";
                  > while($row = mysql_fetch_arr ay($result)) {
                  > $artistID = $row['artistID'];
                  > $email = $row['email'];
                  > $city = $row['city'];
                  > $state = $row['state'];
                  > $country = $row['country'];
                  > echo "<CENTER><HR><F ONT size=\"2\" face=\"Arial, Helvetica,
                  > sans-serif\">
                  > <A HREF=\"javascri pt:;\"
                  > onClick=\"openP rofile('artist. php?artistID=". $artistID.
                  >
                  > "','Artist','re sizable=yes,wid th=600,height=3 00')\">".$first name."
                  > ".$lastname ."</A><BR>"
                  > .$city.", ".$state." ".$country. "</FONT><BR></CENTER>";
                  > }
                  > echo "<HR><CENTE R>If you are already listed above,
                  > congratulations !<BR>
                  > If needed, you may edit your listing using the link at
                  > left.</CENTER>";
                  > mysql_free_resu lt($result);
                  > }
                  >
                  >
                  > Thanx,
                  > Wm
                  >
                  >
                  >[/color]


                    Comment

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